刘明伟回答:网友采纳 an是方程(n^2+n)x^2+(n^2+n-1)x-1=0的根,解得an=1/(n^2+n)或an=-1(舍) 即an=1/[n*(n+1)]=1/n-1/(n+1) Sn=a1+a2+a3+.+an =1-1/2+1/2-1/3+1/3-.+1/n-1/(n+1) =1-1/(n+1) =n/(n+1) limSn=1